Question

我有两个数据框：

df.1 <- data.frame(loc = c('A','B','C','C'), person = c(1,2,3,4), str = c("door / window / table", "window / table / toilet / vase ", "TV / remote / phone / window", "book / vase / car / chair"))

因此，

  loc person                             str
1   A      1           door / window / table
2   B      2 window / table / toilet / vase 
3   C      3    TV / remote / phone / window
4   C      4       book / vase / car / chair

和

df.2 <- data.frame(loc = c('A','B','C'), str = c("book / chair / chair", " table / remote / vase ", "window"))

给出，

  loc                     str
1   A    book / chair / car
2   B  table / remote / vase 
3   C                  window

我想创建一个变量df.1$percentage来计算df.1$str中df.2$str 编辑中元素的百分比，或者：

  loc person                             str percentage
1   A      1           door / window / table       0.00
2   B      2 window / table / toilet / vase        0.50
3   C      3    TV / remote / phone / window       0.25
4   C      4       book / vase / car / chair       0.00

（1有0/3，2有2/4个匹配，3有1/4，4有0/4）

谢谢！

Answer 1

有人可能会提出一个更智能的解决方案，但这是一个简单明了的方法：

library(data.table)
dt1 = data.table(df.1, key = "loc") # set the key to match by loc
dt2 = data.table(df.2)

dt1[, percentage := dt1[dt2][, # merge
           # clean up spaces and convert to strings
           `:=`(str = gsub(" ", "", as.character(str)),
                str.1 = gsub(" ", "", as.character(str.1)))][,
           # calculate the percentage for each row
           lapply(1:.N, function(i) {
                tmp = strsplit(str, "/")[[i]];
                sum(tmp %in% strsplit(str.1, "/")[[i]])/length(tmp)
           })
   ]]

dt1
#   loc person                             str percentage
#1:   A      1           door / window / table          0
#2:   B      2 window / table / toilet / vase         0.5
#3:   C      3    TV / remote / phone / window       0.25
#4:   C      4       book / vase / car / chair          0

Answer 2

您可能知道，data.frame列也可以保存列表（请参阅Create a data.frame where a column is a list）。因此，您可以将str拆分为单词列表：

df.1 <- transform(df.1, words.1 = I(strsplit(as.character(str), " / ")))
df.2 <- transform(df.2, words.2 = I(strsplit(as.character(str), " / ")))

然后合并您的数据：

m <- merge(df.1, df.2, by = "loc")

只需使用mapply计算百分比：

transform(m, percentage = mapply(function(x, y) sum(x%in%y) / length(x),
                                 words.1, words.2))

Answer 3

另一种方式，

test <- data.frame(str1 = df.1[1:nrow(df.2),]$str, str2 = df.2$str)
df.1$percent <- NA
getwords <- function(x) { gsub(" ","",unlist(strsplit(as.character(x),"/"))) }
percent <- function(x,y) {
 sum(!is.na(unlist(sapply(getwords(x), function (d) grep(d, getwords(y))))))/
 length(getwords(x))
}
df.1[1:nrow(df.2),]$percent <- apply(test, 1, function(x) percent(x[1],x[2]))

> df.1
      loc person                             str percent

#   A      1           door / window / table    0.00
#   B      2 window / table / toilet / vase     0.50
#   C      3    TV / remote / phone / window    0.25
#   C      4       book / vase / car / chair      NA

计算两个字符串之间的匹配

3 个答案: